Suppose that a body of an object is experiencing SHM ( simple harmonic motion) and it oscillates with position x(t)=(0.55m)Cos(35s^-1t + pi)
a)What is the period, frequency, and angular frequency
b) calculate its initial velocity and the maximum speed it can have
C) calculate its initial acceleration and the maximum( in magnitude) of the acceleration
d) when the object is in 2.0s , what is its position and location
x = A cos (2 pi f t + pi}
but
cos (a+b) =cos a cos b -sin a sin b
so
x = A[ cos (2pift) cos pi - sin (2 pi f t) sin pi ]
but cos pi = -1 and sin pi = 0
so
x = -A oos 2 pi f t
x = -.55 cos 35 t
2 pi f = 35 = omega
2 pi/T = 35
f = 35/2pi
v = dx/dt = .55(35) sin 35 t
v = 0 at t = 0v = .55*35 max
a = .55* 35^2 cos 35 t
etc :)
To answer these questions, we need to understand the basic equations and definitions related to simple harmonic motion (SHM).
a) Period (T): The time taken for one complete cycle of oscillation.
Frequency (f): The number of complete cycles per unit time.
Angular frequency (ω): The rate at which the object oscillates in radians per unit time.
For the given equation x(t) = (0.55m) * cos(35s^(-1)t + π), we can determine the period, frequency, and angular frequency.
To find the period (T), we use the formula T = 2π/ω, where ω is the angular frequency.
In this case, ω = 35 s^(-1) as given in the equation.
Plugging the value of ω into the formula, we get T = 2π/35 s ≈ 0.18 s.
To find the frequency (f), we use the formula f = 1/T.
Plugging the value of T into the formula, we get f ≈ 1/0.18 s^(-1) ≈ 5.56 Hz.
The angular frequency (ω) is already given as 35 s^(-1).
b) Initial velocity (v₀): The velocity of the object at the start of the oscillation, which can be found using the derivative of the position equation.
Taking the derivative of x(t) with respect to t, we get v(t) = -(0.55m) * 35s^(-1) * sin(35s^(-1)t + π).
To find the initial velocity, we evaluate v(t) at t = 0s.
v₀ = -(0.55m) * 35s^(-1) * sin(π) = -(0.55m) * 35s^(-1) * 0 = 0 m/s.
The maximum speed (v_max) can be found by taking the absolute value of the maximum magnitude of the velocity function.
v_max = |(0.55m) * 35s^(-1)| = 19.25 m/s.
c) Initial acceleration (a₀): The acceleration of the object at the start of the oscillation, which can be found using the derivative of the velocity equation.
Taking the derivative of v(t) with respect to t, we get a(t) = -(0.55m) * 35s^(-1)^2 * cos(35s^(-1)t + π).
To find the initial acceleration, we evaluate a(t) at t = 0s.
a₀ = -(0.55m) * 35s^(-1)^2 * cos(π) = -(0.55m) * 35s^(-1)^2 * (-1) = 19.25 m/s^2.
The maximum magnitude of the acceleration (a_max) occurs when the object is at the extreme points of its oscillation.
a_max = |(0.55m) * 35s^(-1)^2| = 677.75 m/s^2.
d) To find the position (x) and velocity (v) of the object at t = 2.0s, we can substitute t = 2.0s into the equations.
Position at t = 2.0s:
x(2.0s) = (0.55m) * cos(35s^(-1) * 2.0s + π) = (0.55m) * cos(70π + π) = (0.55m) * cos(71π).
To find the location, we can evaluate the value of cos(71π). Note that cos has a period of 2π.
cos(71π) is equal to cos(π), as 71π is equivalent to 1π after removing the multiple of 2π.
cos(π) = -1.
Therefore, x(2.0s) = (0.55m) * (-1) = -0.55m.
Velocity at t = 2.0s:
v(2.0s) = -(0.55m) * 35s^(-1) * sin(35s^(-1) * 2.0s + π) = -(0.55m) * 35s^(-1) * sin(70π + π).
Similarly to the position calculation, sin(70π + π) is equivalent to sin(π), which is 0.
Therefore, v(2.0s) = -(0.55m) * 35s^(-1) * 0 = 0 m/s.
Hence, when the object is at t = 2.0s, its position is x = -0.55m and its velocity is v = 0 m/s.